Holooly Plus Logo

Question 5.5: Estimate the subsonic parasite-drag coefficient This example......

Estimate the subsonic parasite-drag coefficient
This example will show how the subsonic parasite-drag coefficient for the
F-16 can be estimated at a specific altitude. It is important to note that, unlike other aerodynamic coefficients, the subsonic drag coefficient is a function of altitude and Mach number, since the Reynolds number over the surfaces of the aircraft will vary with altitude and Mach. This example will assume that the aircraft is flying at an altitude of 30,000 ft and has a Mach number of 0.4 (to match available flight test data). The theoretical wing area of the F-16 is 300 ft², which will serve as the reference area for the aircraft: S_{\mathrm{ref}}=300\;\mathrm{ft}^{2}

Step-by-Step
The 'Blue Check Mark' means that this solution was answered by an expert.
Learn more on how do we answer questions.

The first task is to estimate the wetted area of the various surfaces of the aircraft. A good estimate of these areas has been completed in Brandt et al.  (2004) and is reproduced here in Fig. 5.24 and Tables 5.2 and 5.3.The aircraft is approximated with a series of wing-like and fuselage-like shapes, as shown in Fig. 5.24. The wetted area for these simplified geometric surfaces is approximated using equations (5.38 and 5.40).

S_{\mathrm{wet}}\approx2.0(1\ +\ 0.2t/c)S_{\mathrm{exposed}} (5.38)

\begin{array}{l}{{S_{\mathrm{wet_{nose}}}=0.75\pi D L_{\mathrm{nose}}}}\\ {{S_{\mathrm{wet_{body}}}=\pi D L_{\mathrm{body}}}}\\ {{S_{\mathrm{wet_{tail}}}=0.72\pi D L_{\mathrm{tail}}}}\end{array} (5.40)

This simplified approximation yields a total wetted area of 1418 ft², which is very close to the actual wetted area of the F-16, which is 1495 ft² [Brandt et al.(2004)]. Now that these wetted areas have been obtained, the parasite-drag coefficient for each surface can be estimated

Wing. First, estimate the mean aerodynamic chord of the wing as

{\mathrm{m.a.c.}}={\frac{2}{3}}{\bigg(}c_{r}+c_{t}-{\frac{c_{r}c_{t}}{c_{r}+c_{t}}}{\bigg)}={\frac{2}{3}}{\bigg(}14+3.5-{\frac{(14)(3.5)}{14+3.5}}{\bigg)}=9.800\,{\mathrm{ft}}

and use the m.a.c. to calculate the Reynolds number for the equivalent rectangular wing as

\mathrm{Re}_{L}={\frac{\rho_{\infty}U_{\infty}\mathrm{m.a.c.}}{\mu_{\infty}}}={\frac{(0.000891~\mathrm{slug/ft}^{3})(397.92~\mathrm{ft}/s)(9.800~\mathrm{ft})}{3.107\times10^{-7}\mathrm{lb}-s/\mathrm{ft}^{2}}}

 

={11.18}\times10^{6}

Finally, the total skin-friction coefficient for the wing is calculated as

\bar{C}_{f}=\frac{0.455}{(\log_{10}\mathrm{Re}_{L})^{258}}-\frac{1700}{\mathrm{Re}_{L}}=\frac{\mathrm{0.455}}{(\log_{10}1.18\times10^{6})^{258}}-\frac{1700}{11.18\times10^{6}}

= 0.00280
Assume for the sake of this example that the wing is aerodynamically smooth, so no roughness correction will be applied. However, a form factor correction should be performed. From Fig. 5.20, for a leading edge sweep of 40° and a thickness ratio of 0.04, K = 1.06 and the parasite drag coefficient is

C_{D_{0}}=\frac{K\bar{C}_{f}S_{\mathrm{wet}}}{S_{\mathrm{ref}}}=\frac{(1.06)(0.00280)(419.4\,\mathrm{ft}^{2})}{300\,\mathrm{ft}^{2}}=0.00415

The other wingLlike components of the F-16 have had similar analysis performed, resulting in the zero-lift drag predictions presented in Table 5.4.

Fuseiage.

L=L_{\mathrm{nose}}+L_{\mathrm{fuselage}}+L_{\mathrm{boattail}}=6+39+4=49.0\,\mathrm{ft}

 

\mathrm{Re}_{L}={\frac{(0.000891\,{\mathrm{slug/ft^{3}}})(397\,92\,{\mathrm{ft}}/{\mathrm{s}})(49.0\,{\mathrm{ft}})}{3.107\times10^{-7}\,{\mathrm{lb}}\,-s/{\mathrm{ft}}^{2}}}=55.91\times10^{6}

 

\overline{{{C}}}_{f}=\frac{0.455}{(\mathrm{log}_{10}\,55\,.91\times10^{6})^{2.58}}-\frac{1700}{55.91\times10^{6}}=0.00228

From Fig 5.22, for a fineness ratio of L/D = 49.0/0.5*(2.5 + 5.0) = 13.067, K = 1.05 (requires some extrapolation from the graph) and the parasite drag coefficient is

C_{D_{0}}=\frac{K\bar{C}_{f}S_{\mathrm{wet}}}{S_{\mathrm{ref}}}=\frac{(1.05)(0.00228)(656.5\,\mathrm{t}^{2})}{300\,\mathrm{ft}^{2}}=0.00524

The other fuselage-like components of the F-16 have had similar analysis performed, resulting in the zero-lift drag predictions presented in Table 5.5. The total aircraft zero-lift drag coefficient (assuming aerodynamically smooth surfaces) is

C_{D_{0}}=C_{D_{0}\mathrm{\,(Wings)}}+C_{D_{0}\mathrm{\,(Fuselages )}}=0.00710+0.00590=0.01300

Since the total wetted-area estimate from this analysis was 5.4% lower than the actual wetted area of the F-16 (something which could be improved with a better representation of the aircraft surfaces, such as from a CAD geometry), it would be reasonable to increase the zero-lift drag value by 5.4% to take into account the simplicity of the geometry model. This would result in a zero-lift drag coefficient of C_{D_{0}}=0.01370.

Again, this result assumes that the surfaces are aerodynamically smooth, that there are drag increments due to excrescence or base drag, and that there is no interference among the various components of the aircraft. If we assume that the other components of drag account for an additional 10%, then our final estimate for the zero-lift drag coefficient would be C_{D_{0}}=0.0151. Initial flight test data for the F-16 showed that the subsonic zero-lift drag coefficient varied between C_{D_{0}}=0.0160 and C_{D_{0}} = 0.0190 after correcting for engine effects and the presence of missiles in the flight test data [Webb, et al. (1977)].These results should be considered quite good for a fairly straightforward method that can be used easily on a spreadsheet. 

TABLE 5.2 F-16 Wing-Like Surface Wetted Area Estimations
Surface Span, ft c_{r},ft c_{t},ft t/c S_{\mathrm{{wet}}},\,{ft}^{2}
Wing (1 and 2) ’12 14 3.5 0.04 419.4
Horizontal tail (3 and 4) 6 7.8 2 0.04 117.5
Strake (5 and 6) 2 9.6 0 0.06 38.6
Inboard vertical tail (7) 1.4 12.5 6 0.10 26.3
Outboard vertical tail (8) 7 8 3 0.06 77.3
Dorsal fins (9 and 10)* 1.5 5 3 0.03 23.9

*not shown in Fig. 5.24

TABLE 5,3 F-16 Fuselage-Like Surface Wetted Area Estimations
Surface Length, ft Height, ft width, ft S_{\mathrm{{wet}}},\,{ft}^{2} Net S_{\mathrm{{wet}}},\,{ft}^{2}
Fuselage (cylinder 1) 39 2.5 5 551.3 551.3
Nose (cone 1) 6 2.5 5 42.4 42.4
Boattail (cylinder 2) 4 6 6 62.8 62.8
Side (half cylinder 1 and 2) 24 0.8 1 67.9 29.5
Canopy(halfcylinder3) 5 2 2 15.7 5.7
Engine (half cylinder4) 30 2.5 5 180 32.1
Canopyfront(half cone 1) 2 2 2 3.1 1.1
Canopy rear (half cone 2) 4 2 2 6.3 2.3
TABLE 5.4 F-16 Wing-Like Surface Zero-Lift Drag Estimations
Surface m.a.c., ft {R e}_{L}(\times10^{-6}) {\overline{C}}_{f} K C_{D_{0}}
Wing (1 and 2) 9.800 11.18 0.0028 1.06 0.00415
Horjzontal tail (3 and 4) 5.472 6.568 0.00296 1.06 0.00123
Strake (5 and 6) 6.400 7.303 0.00293 1.04 0.00039
Inboard vertical tail (7) 9.631 10.99 0.00280 1.04 0.00026
Outboard vertical tail (8) 5.879 6.708 0.00295 1.08 0.00082
Dorsal Fins (9 and 4.083 4.660 0.00304 1.04 0.00025
Total 0.00710
TABLE 5.5 F-16 Fuselage-Like Surface Zero-Lift Drag Estimations
Surface Length, ft \mathrm{Re}_{L}(\times10^{-6}) {\overline{C}}_{f} k C_{D_{0}}
Fuselage + nose + boattail 49.0 55.91 0.00228 1.05 0.00524
Side (half cylinder 1 and 2) 24.0 27.39 0.00251 1.01 0.00025
Canopy (front + center + rear) 11.0 12.55 0.00276 1.25 0.00011
Engine (half cylinder 4) 30.0 34.23 0.00244 1.15 0.00030
Total 0.0059
Screenshot 2023-06-22 142419
Screenshot 2023-06-22 143859
Screenshot 2023-06-22 144929

Related Answered Questions